Let ∆ABC is a triangle such that side length of AB is c , BC is a and CA is b .
Given, AB = c = 8
m∠A=60°
m∠C=45°
m∠B = (180 - 45 - 60)° = 75°
use sine rule to get b and c
\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}asinA=bsinB=csinC
so, \frac{sin60^{\circ}}{8}=\frac{sin45^{\circ}}{c}=\frac{sin75^{\circ}}{b}8sin60∘=csin45∘=bsin75∘
sin60°/8 = sin45°/c
(√3/2)/8 = (1/√2)/c
√3/16 = 1/√2c => c = 16/√6
also, sin60°/8 = sin75°/b
b = 8sin75°/sin60°
= {8 × (√3 + 1)/2√2}/{√3/2}
= 4√2(√3 + 1)/√3
hence, perimeter of ∆ABC = a + b + c
= 8 + 16/√6 + 4√2(√3 + 1)/√3
= 8 + (16 + 8√3 + 8)/√6
= 8 + (24 + 8√3)/√6
= 8 + 4√6 + 4√2
area of ∆ABC = 1/2 absinC
= 1/2 × 8 × 4√2(√3 + 1)/√3 × sin45°
= 4 × 4√2(√3 + 1)/√3 × 1/√2
= 16(√3 + 1)/√3
= (48 + 16√3)/3
= 16 + 16/√3
